According to the question it is given that,

Radius of the cylindrical portion of the rocket (R) $=2.5m$

Height of the cylindrical portion of the rocket (H) $=21m$

Slant Height of the Conical surface of the rocket (L) $=8m$

Curved Surface Area of the Cone $\left( {{S}_{1}} \right)=\pi rL=\pi \left( 2.5 \right)\left( 8 \right)=20\pi $

Then,

Curved Surface Area of the Cone $\left( {{S}_{2}} \right)=2\pi RH+\pi {{R}^{2}}$

${{S}_{2}}=\left( 2\pi \times 2.5\times 21 \right)+\pi {{\left( 2.5 \right)}^{2}}$

${{S}_{2}}=\left( \pi \times 105 \right)+\left( \pi \times 6.25 \right)$

Thus, the total curved surface area S is

$S={{S}_{1}}+{{S}_{2}}$

$S=\left( \pi 20 \right)+\left( \pi 105 \right)+\left( \pi 6.25 \right)$

$S=\left( 22/7 \right)\left( 20+105+6.25 \right)=22/7\times 131.25$

$S=412.5{{m}^{2}}$

Now, the total Surface Area of the Conical Surface $=412.5{{m}^{2}}$

Then, calculating the volume of the rocket

Volume of the conical part of the rocket $\left( {{V}_{1}} \right)=1/3\times 22/7\times {{R}^{2}}\times h$

${{V}_{1}}=1/3\times 22/7\times {{\left( 2.5 \right)}^{2}}\times h$

Assume, h be the height of the conical portion in the rocket.

We all know that,

${{L}^{2}}={{R}^{2}}+{{h}^{2}}$

${{h}^{2}}={{L}^{2}}-{{R}^{2}}={{8}^{2}}-{{2.5}^{2}}$

$h=7.6m$

Using the value of h, we will get

Volume of the conical part $\left( {{V}_{1}} \right)=1/3\times 22/7\times {{2.5}^{2}}\times 7.6{{m}^{2}}=49.67{{m}^{2}}$

Then,

Volume of the Cylindrical Portion $\left( {{V}_{2}} \right)=\pi {{R}^{2}}h$

${{V}_{2}}=22/7\times {{2.5}^{2}}\times 21=412.5{{m}^{2}}$

Thus, the total volume of the rocket $={{V}_{1}}+{{V}_{2}}$

$V=412.5+49.67=462.17{{m}^{2}}$

Hence, the total volume of the Rocket is $462.17{{m}^{2}}$